Construction of Segment on Given Circle Admitting a Given Angle

Theorem

From any given circle, it is possible to cut off a segment which admits an angle equal to any given rectilineal angle.

Construction

Let $ABC$ be the given circle, and let $D$ be the given rectilineal angle.

Let $EF$ be drawn tangent to $ABC$ at $B$.

Let $\angle FBC$ be constructed equal to $\angle D$.

Then $BC$ is a straight line which forms the base of a segment which admits an angle equal to $D$.

Proof

Let $A$ be selected anywhere on the circle opposite the segment in question.

From Angles made by Chord with Tangent‎, $\angle BAC = \angle FBC$.

But $\angle FBC = \angle D$.

Hence the result.

$\blacksquare$

Historical Note

This is Proposition 34 of Book III of Euclid's The Elements.